Convergence of series using subsequence. If $\sum^\infty_\mathrm{n=1}a_n$ is a convergent series of positive numbers, and $\{a_{n_i}\}^\infty_\mathrm{i=1}$ is a subsequence of $\{a_{n}\}^\infty_\mathrm{n=1}$,
prove that $\sum^\infty_\mathrm{i=1}a_{n_i}$ converges.
I see several places on this site that talk about the convergence of sequences in this context. Specifically, I'm confused why $\sum^\infty_\mathrm{i=1}a_{n_i}$ , a series , must converge given the above information. If this is a simple proof, then that would be great. If the proof relies on some comparison test, what is the logic being used?
 A: Note that $\sum a_{n_i}\leq\sum a_n$, because you are adding only some of the terms of $a_n$.
A: Denote by $s_j$ the $j$-th partial sum of $a_{n_i}$ and denote by $t_N$ the $N$-th partial sum of $a_n$. Choose $N(j)$ so that
$$
s_j \le t_{N(j)}
$$
That is, choose $N$ large enough so that all the terms in $s_j$ also appear in $t_{N(j)}$. Since $t_N$ converges, the subsequence $t_{N(j)}$ converges also. Thus passing to the limit as $j \to \infty$,
$$
\sum_{i=1}^{\infty} a_{n_i} = \lim_{j\to\infty} s_{j} \le \sum_{n=1}^{\infty} a_n < \infty
$$
A: By Cauchy criterion, there exists $N \in \mathbf{N}$ such that for all $n > m \geq N$, we have
$$|a_m + a_{m + 1} + \ldots + a_{n}| < \epsilon \tag{1}$$
Let $(a_{n_i})$ be an arbitrary subsequence of $(a_n)$.
$$(a_{n_i}) := (a_{n_1},a_{n_2},\ldots)$$
Let $a_{n_j}$ and $a_{n_l}$ be any two terms of the subsequence satisfying $n_l > n_j \geq N$. Then, from (1), it follows that, for all $n_l > n_j \geq N$, we have
$$|a_{n_j} + a_{{n_j}+1} + \ldots + a_{n_l}| < \epsilon$$
Since all numbers on the LHS are positive, we can only leave the terms of the subsequence and drop the remaining terms and write for all $n_l > n_j \geq N$:
$$|a_{n_{j}} + a_{n_{j+1}} + \ldots + a_{n_l}| < |a_{n_j} + a_{n_{j}+1} + \ldots + a_{n_l}| < \epsilon$$
Consequently, by the Cauchy Criterion, $\sum_{i=1}^{\infty}a_{n_i}$ is convergent.
